Acceleration & Mass: Effects on Weight

[edguy99]

1/ I suspect most would agree with this statement:

"If you were a 1kg box, stationary against a spring, the spring is released and you were accelerated to 95% the speed of light, you would weigh considerably more then 1kg"

Ie. If a distant observer were watching you, you would hit other objects with what appears to be considerably more then 1kg of mass.

2/ I am unsure the correctness of this:

"If you were a 1kg box, stationary above a very large gravitational mass (but very small size ie. black hole), and started to drop. When you reached 95% of the speed of light, you would weigh considerably LESS then 1 kg"

... or should this be more?

 

[JesseM]

1/ I suspect most would agree with this statement:

"If you were a 1kg box, stationary against a spring, the spring is released and you were accelerated to 95% the speed of light, you would weigh considerably more then 1kg"

Ie. If a distant observer were watching you, you would hit other objects with what appears to be considerably more then 1kg of mass.

What does it mean to measure mass in terms of what happens when you "hit other objects"? I suppose you could look at the amount of kinetic energy it hits the object with, but the kinetic energy of an object moving at any velocity is $$(\gamma - 1) mc^2$$ where m is the object's rest mass, in this case 1kg.

2/ I am unsure the correctness of this:

"If you were a 1kg box, stationary above a very large gravitational mass (but very small size ie. black hole), and started to drop. When you reached 95% of the speed of light, you would weigh considerably LESS then 1 kg"

Again what method are you using to "weigh" the box?

 

[pervect]

1/ I suspect most would agree with this statement:

"If you were a 1kg box, stationary against a spring, the spring is released and you were accelerated to 95% the speed of light, you would weigh considerably more then 1kg"

If that's a question, it's a funny way to word it. If it's a statement, it seems unsupported.

 

[Bill_K]

What does it mean to measure mass in terms of what happens when you "hit other objects"?

I believe he's talking about momentum.

Let me rephrase his second question. You can determine the inertial mass of an object by performing F=ma experiments, that is, apply a known force and observe the acceleration that results. In general relativistic terms this is a local experiment. I believe his is question is, if you do the experiment once in a flat region of space and once deep in the gravitational field of a massive body will you get the same answer.

 

[edguy99]

That was a thought experiement where I guess some clarification is needed. Let me start with an easier setup before getting back to the original question, perhaps this will put the question in better context.

If a particle is accelerated to 95% of the speed of light in an acceleration device and then crashed into another particle, what is used in calculating the paths seen in a cloud chamber? Clearly the speed of both particles are known (one at 95% of c and the other is stationary) and the mass of the resting particle is known. Is the original rest mass of the moving particle used or the "heavier" mass used? Would the same be applicable to both of the thought experiments in the original post or is there a fundamental difference between being accelerated in a particle accelerator and falling towards a black hole?

 

[jtbell]

If a particle is accelerated to 95% of the speed of light in an acceleration device and then crashed into another particle, what is used in calculating the paths seen in a cloud chamber?

Conservation of (total) momentum and conservation of (total) energy. Assuming for the sake of simplicity that the particles are traveling along a single straight line (e.g. the x-axis) before and after the collision, but allowing the the particles to "transmute" in the collision so they have different rest-masses before and after the collision:

$$A + B \rightarrow C + D$$

$$\frac{m_{0A} v_A}{\sqrt{1 - v_A^2 / c^2}} + \frac{m_{0B} v_B}{\sqrt{1 - v_B^2 / c^2}} = \frac{m_{0C} v_C}{\sqrt{1 - v_C^2 / c^2}} + \frac{m_{0D} v_D}{\sqrt{1 - v_D^2 / c^2}}$$

$$\frac{m_{0A} c^2}{\sqrt{1 - v_A^2 / c^2}} + \frac{m_{0B} c^2}{\sqrt{1 - v_B^2 / c^2}} = \frac{m_{0C} c^2}{\sqrt{1 - v_C^2 / c^2}} + \frac{m_{0D} c^2}{\sqrt{1 - v_D^2 / c^2}}$$

If you know the four rest-masses, and the initial velocities, you have two equations for two unknowns (the final velocities).

 

[ZealScience]

1/ I suspect most would agree with this statement:

"If you were a 1kg box, stationary against a spring, the spring is released and you were accelerated to 95% the speed of light, you would weigh considerably more then 1kg"

Ie. If a distant observer were watching you, you would hit other objects with what appears to be considerably more then 1kg of mass.

2/ I am unsure the correctness of this:

"If you were a 1kg box, stationary above a very large gravitational mass (but very small size ie. black hole), and started to drop. When you reached 95% of the speed of light, you would weigh considerably LESS then 1 kg"

... or should this be more?

Are you talking special relativity here? If you are the reference frame, I think at least what you observe is that mass is greater than 1kg. Because kinetic energy has considerably increased. Then the equivalent mass would increase a lot.

And what do you mean by the second? Mass depends on the energy that the object has. If there is no change in energy in that reference frame, mass won't change. Sounds like you are talking about weight rather than energy, I mean how to weigh the mass?

 

[edguy99]

Conservation of (total) momentum and conservation of (total) energy. Assuming for the sake of simplicity that the particles are traveling along a single straight line (e.g. the x-axis) before and after the collision, but allowing the the particles to "transmute" in the collision so they have different rest-masses before and after the collision:

... If you know the four rest-masses, and the initial velocities, you have two equations for two unknowns (the final velocities).

Thank you, that was very helpful (although it took me a while!). Consider a 1kg MassB crashed into a 1kg MassA at very high speeds. Assume they "stick" together after the crash as MassC and we keep MassD=0 to keep it simple.

Initial conditions (speed of light c = 3x10^8 m/sec):
MassA=1, SpeedA=0, MassB=1, SpeedB=50%c ==>> SpeedC=0.8x10^8, MassC=2.08
MassA=1, SpeedA=0, MassB=1, SpeedB=90%c ==>> SpeedC=1.9x10^8, MassC=2.57
MassA=1, SpeedA=0, MassB=1, SpeedB=95%c ==>> SpeedC=2.2x10^8, MassC=2.84
MassA=1, SpeedA=0, MassB=1, SpeedB=98%c ==>> SpeedC=2.5x10^8, MassC=3.48
MassA=1, SpeedA=0, MassB=1, SpeedB=99%c ==>> SpeedC=2.6x10^8, MassC=4.02

------------------
From the original post:

1/ I suspect most would agree with this statement:

"If you were a 1kg box, stationary against a spring, the spring is released and you were accelerated to 95% the speed of light, you would weigh considerably more then 1kg"

Ie. If a distant observer were watching you, you would hit other objects with what appears to be considerably more then 1kg of mass.

The calculations seem to support the idea that high speed particles hit other particles as if they were "heavier". I assume this is observed when protons and neutrons are accelerated to high speeds where the energy of magnets has been converted not only to the speed of the particles, but also a heavier "effective" mass.

I have seen this quote http://www.mathpages.com/rr/rrtoc.htm" :

In Einstein's second paper on relativity in 1905, he explicitly concludes "Radiation carries inertia between emitting and absorbing bodies". It is important that not only does something receive a "kick" from the momentum of the energy, but the internal inertia (i.e., the inertial mass) of the body is actually increased.

 

[PAllen]

Thank you, that was very helpful (although it took me a while!). Consider a 1kg MassB crashed into a 1kg MassA at very high speeds. Assume they "stick" together after the crash as MassC and we keep MassD=0 to keep it simple.

Initial conditions (speed of light c = 3x10^8 m/sec):
MassA=1, SpeedA=0, MassB=1, SpeedB=50%c ==>> SpeedC=0.8x10^8, MassC=2.08
MassA=1, SpeedA=0, MassB=1, SpeedB=90%c ==>> SpeedC=1.9x10^8, MassC=2.57
MassA=1, SpeedA=0, MassB=1, SpeedB=95%c ==>> SpeedC=2.2x10^8, MassC=2.84
MassA=1, SpeedA=0, MassB=1, SpeedB=98%c ==>> SpeedC=2.5x10^8, MassC=3.48
MassA=1, SpeedA=0, MassB=1, SpeedB=99%c ==>> SpeedC=2.6x10^8, MassC=4.02

------------------
From the original post:


The calculations seem to support the idea that high speed particles hit other particles as if they were "heavier". I assume this is observed when protons and neutrons are accelerated to high speeds where the energy of magnets has been converted not only to the speed of the particles, but also a heavier "effective" mass.

I have seen this quote http://www.mathpages.com/rr/rrtoc.htm" :

In Einstein's second paper on relativity in 1905, he explicitly concludes "Radiation carries inertia between emitting and absorbing bodies". It is important that not only does something receive a "kick" from the momentum of the energy, but the internal inertia (i.e., the inertial mass) of the body is actually increased.

Since you'r having them stick together, the excess mass of the final object is kinetic energy of A absorbed into the final object. It is just a matter of interpretation, but I would say the fast moving object has high momentum and kinetic energy rather than high mass. Then, the high fused object mass comes from conversion of kinetic energy to internal energy (leading to mass).

 

[pervect]

Mass turns out to be a trickier concept than it first appears. While it overstates the case slightly, http://adsabs.harvard.edu/abs/2006PhTea..44...40H, "There is no really good definition of mass" is an example in the literature of some of the problems in defining mass.

In General Relativity, for instance, there are several definitions of mass - it's a symptom of the above mentioned problem, but in my opinion it's a bit less controversial to say that there are "too many" definitions of mass rather than to imply that there are none. (It is true that none of the several commonly used definitions is completely general, but that gets outside the scope of this post).

Arguments about "mass" tend to quickly become semantic, about the meaning of the word, and generally unproductive.

Special relativity defines two different sorts of mass - neither of which can really be considered to be the source of gravity in GR - the relativistic mass, and the invariant mass. Everyone agrees that relativistic mass is a synonym for energy. Many people, including myself, believe that since relativistic mass is equivalent to energy, it's better to call it energy than relativistic mass. The other sort of mass commonly used in special relativity, invariant mass, is a property of object independent of the observer. The energy of an object does depend on the observer (i.e. if the object is moving fast, or not at all). This is another advantage of the invariant mass, it's a property of an object independent of the observer, while the relativistic mass and or energy is not.

The sorts of mass associated most naturally with gravity, for instance the Komar, ADM, and Bondi masses are not defined in the same manner as mass is in special relativity, and, while they are interesting, are somewhat difficult to talk about in lay terms.

 

[edguy99]

Mass turns out to be a trickier concept than it first appears. While it overstates the case slightly, http://adsabs.harvard.edu/abs/2006PhTea..44...40H, "There is no really good definition of mass" is an example ...

...The sorts of mass associated most naturally with gravity, for instance the Komar, ADM, and Bondi masses are not defined in the same manner as mass is in special relativity, and, while they are interesting, are somewhat difficult to talk about in lay terms.

I was unable to gain access to your link, but I have looked up and read some of these references, thank you. I am not sure what they predict for:

2/ "If you were a 1kg box, stationary above a very large gravitational mass (but very small size ie. black hole), and started to drop. When you reached 95% of the speed of light, you would weigh considerably LESS then 1 kg"

The concept of weight here is: if you hit something, how big and fast moving is the result of your collision.

I would like to suggest that the particle losses weight as it falls in a gravity field. If you consider the particle accelerator, there is a source of energy that supplies the additional mass that is eventally felt. Under gravity, if the 1 kg particle goes from zero speed to 95% of the speed of light, the "total mass + energy" of the system cannot be changed so the additional energy must have come from the mass of the system, suggesting that the 1 kg particle now may only weigh say 0.2 kg (not calculated, just a guess?)

Is is correct to say that a particle loses inertial mass as it falls in a gravity field, the same way as we speak of a particle gaining inertial mass as it is accelerated in an accelerator?

 

[pervect]

The concept of weight here is: if you hit something, how big and fast moving is the result of your collision.

You can answer those questions about 'hitting something' if you know the momentum and energy of the impacting object.

If your system is reasonably small and isolated, you can use the special relativity concepts of momentum and energy. This is the same general approach as you would use to solve a collision problem in Newtonian mechanics, except that the formula for the conserved momentum is different. It would be an unfortunate error to insist on using the Newtonian formula for momentum where they don't apply, just because you remember them from school and didn't study relativity. So, just use the correct relativistic formula for momentum (look them up if you need to!), and you're all set. Your equations are simple : the momentum before the collision is the same as the momentum after the collision.

One more wrinkle that will make actual collisions a bit complex, is that neither the ideal of an inelastic collision nor the ideal of an elastic collision will be likely to be 100% accurate, especially for relativistic collisons.

An ideal elastic collision will not convert any energy of motion into heat, an ideal inelastic collision will have both bodies "stick together", and won't generate an expanding exploding debris cloud, or a shower of exotic particles, both of which will undoubtedly happen in a relativistic collision as some of the energy of motion gets transformed into heat and other forms of energy.

If your system is not reasonably isolated, so it can't be modeled as non-interacting objects (there are some fields involved), or if it is very big (operationally, you can tell if it's big if the Newtonian estimate of it's gravitational self-energy is a significant part of the system's total energy), you'll need more information and a more sophisticated approach to handle collisions.

So to sum it up - you keep asking about "mass", but when we ask why, it appears that you are asking about problems you don't need to know the "mass" to solve.

 

[edguy99]

You can answer those questions about 'hitting something' if you know the momentum and energy of the impacting object.

If your system is reasonably small and isolated, you can use the special relativity concepts of momentum and energy. This is the same general approach as you would use to solve a collision problem in Newtonian mechanics, except that the formula for the conserved momentum is different. It would be an unfortunate error to insist on using the Newtonian formula for momentum where they don't apply, just because you remember them from school and didn't study relativity. So, just use the correct relativistic formula for momentum (look them up if you need to!), and you're all set. Your equations are simple : the momentum before the collision is the same as the momentum after the collision.

One more wrinkle that will make actual collisions a bit complex, is that neither the ideal of an inelastic collision nor the ideal of an elastic collision will be likely to be 100% accurate, especially for relativistic collisons.

An ideal elastic collision will not convert any energy of motion into heat, an ideal inelastic collision will have both bodies "stick together", and won't generate an expanding exploding debris cloud, or a shower of exotic particles, both of which will undoubtedly happen in a relativistic collision as some of the energy of motion gets transformed into heat and other forms of energy.

If your system is not reasonably isolated, so it can't be modeled as non-interacting objects (there are some fields involved), or if it is very big (operationally, you can tell if it's big if the Newtonian estimate of it's gravitational self-energy is a significant part of the system's total energy), you'll need more information and a more sophisticated approach to handle collisions.

So to sum it up - you keep asking about "mass", but when we ask why, it appears that you are asking about problems you don't need to know the "mass" to solve.

Sorry, you lost me. The question is:

Is it correct to say that a particle loses inertial mass as it falls in a gravity field, the same way as we speak of a particle gaining inertial mass as it is accelerated in an accelerator?

To rephrase: Do you think a particle gains or loses (or stay the same) with respect to its inertial mass as it falls into a gravity field?

 

[harrylin]

Still not sure from the way you phrase it and it certainly depends on definitons, but maybe this helps:
- When you accelerate something with a particle accelerator, you increase its kinetic energy and thus also its relativistic mass
- When you consider a closed system of two objects falling towards each other, the total energy is conserved. Thus the gain in kinetic energy (which may be transformed into radiation) is at the cost of potential energy.

 

[edguy99]

Still not sure from the way you phrase it and it certainly depends on definitons, but maybe this helps:
- When you accelerate something with a particle accelerator, you increase its kinetic energy and thus also its relativistic mass
- When you consider a closed system of two objects falling towards each other, the total energy is conserved. Thus the gain in kinetic energy (which may be transformed into radiation) is at the cost of potential energy.

I guess if the potential energy is viewed as part of the relativistic mass then perhaps this is the same thing. How much potential energy could eventually be drawn? Ie. can the value mc^2 be reached?

 

[pervect]

Sorry, you lost me. The question is:

Is it correct to say that a particle loses inertial mass as it falls in a gravity field, the same way as we speak of a particle gaining inertial mass as it is accelerated in an accelerator?

To rephrase: Do you think a particle gains or loses (or stay the same) with respect to its inertial mass as it falls into a gravity field?

The short polite answer that it is that what you said is at best ambiguous and confusing, at least in the context of relativity, because you are using a Newtonian term, "inertial mass", that doesn't translate well to relativity.

The followup would be to ask you how you intend to measure this "inertial mass", what experiment you'd intend to perform to show that the particle "loses inertial mass".

A not-very-polite but generally accurate way of talking about statements that can't be decided on the base of experiment is that "they are not even wrong". Because they aren't stated clearly enough to be disproven.

Sorry that you couldn't read the reference I supplied, there are some other useful references out there, Max Jammer has a couple of books, for instance. "Concepts of Mass in Classical and Modern Physics" and "Concepts of Mass in Contemporary Physics and Philosophy".

I would suggest not speaking of a particle gaining "inertial mass" when it is accelerated in an accelerator. Instead, say it gains energy.

When you start talking about what happens to the mass or energy a particle when you lower it into a gravitational field, you move into the realm of GR. Right where the hard stuff I mentioned is :-(.

The very short version of what happens to the energy of a particle when you lower it into a gravitational field is that it does, in fact, go down, in those cases where we have an adaquate definition of energy in GR (which is not always!.) So, what one might say is that the energy-at-infinity of the particle goes down. Note that there are some useful ways of talking about energy that do NOT go down, so some care is needed here.

Another thing you can say that may be similar to what you appear to be thinking is that if you have a 1kg object "at infinity", and you drop it into a black hole in such a manner that you extract energy from it (say lowering it with a long rope, though that's not practical), that the black hole gains less than 1kg of mass.

[add]
While I'm pointing out cases where things similar to what you say could be correct from experiment, I should also point out different circumstances where what I think you're attempting to say would disagree with experiment. For instance, if you have a pan balance, and you weigh a 1kg object with the pan balance, you'll get 1kg, no matter whether you are deep in a gravity well, or far away from any massive bodies.

If you have a calibrated spring that puts out one Newton of force, and you measure the acceleration of a 1 kg body with local clocks, you'll always get 1 m/s^2, regardless of your position in a gravity well.

 

[edguy99]

... I would suggest not speaking of a particle gaining "inertial mass" when it is accelerated in an accelerator. Instead, say it gains energy ...

Using the calculations supplied by jtbell, involving conservation of total energy and conservation of total momentum, I calculated the following for a particle accelerated and crashed into another particle (both start at 1 kg and one is sped up to 95%c and crashed into the other):

MassA=1, SpeedA=0, MassB=1, SpeedB=95%c ==>> SpeedC=2.2x10^8, MassC=2.84


I write computer animations of this type of thing and this appears what should be used to properly model this type of acceleration and crash. I am having problems modeling an object falling into a black hole, hence the question.

If that same particle was accelerated in a gravitational field to 95%c and hit a non-moving particle would the resulting mass be 2.84? At first I thought the above calculation would be correct. Now I have serious doubts as the overall energy and momentum of the system would not be correct. The only way it looks correct is if MassC ends up less then 2? I have tried to be very precise in the formulation of the problem. A 1 kg mass is falling into a black hole, when it reaches the speed of 95%c, it hits and sticks to a 1 kg stationary mass. Is the resulting mass greater or less then 2 kg and what speed is it going? I ask the question as I do not know the answer. Hopefully, it is not "unsolvable" (perhaps it is...).

 

[PAllen]

Using the calculations supplied by jtbell, involving conservation of total energy and conservation of total momentum, I calculated the following for a particle accelerated and crashed into another particle (both start at 1 kg and one is sped up to 95%c and crashed into the other):

MassA=1, SpeedA=0, MassB=1, SpeedB=95%c ==>> SpeedC=2.2x10^8, MassC=2.84


I write computer animations of this type of thing and this appears what should be used to properly model this type of acceleration and crash. I am having problems modeling an object falling into a black hole, hence the question.

If that same particle was accelerated in a gravitational field to 95%c and hit a non-moving particle would the resulting mass be 2.84? At first I thought the above calculation would be correct. Now I have serious doubts as the overall energy and momentum of the system would not be correct. The only way it looks correct is if MassC ends up less then 2? I have tried to be very precise in the formulation of the problem. A 1 kg mass is falling into a black hole, when it reaches the speed of 95%c, it hits and sticks to a 1 kg stationary mass. Is the resulting mass greater or less then 2 kg and what speed is it going? I ask the question as I do not know the answer. Hopefully, it is not "unsolvable" (perhaps it is...).

It depends on who is measuring it. If you are measuring local to the initial stationay mass at the moment of collition, and the infalling mass is locally at 95%c, then the result would be the same as in deep space. This a fundamental principle of GR: everywhere is locally Minkowski space.

 

[edguy99]

It depends on who is measuring it. If you are measuring local to the initial stationay mass at the moment of collition, and the infalling mass is locally at 95%c, then the result would be the same as in deep space. This a fundamental principle of GR: everywhere is locally Minkowski space.

I agree if the infalling mass is still 1 kg, but after falling from a speed of 0 to 95%c, would the infalling mass still be 1 kg if you are measuring local to the initial stationary mass at the moment of collision?

The question I am really trying to get to: What does the distant observer see as the apparent speed and mass of the resulting object from the collision. The distant observer will see a certain speed and can infer a mass based on additional collisions. You see lots of correct animations for particle colliders, but you don't see many gravity animations of this since the computer programmers like me (maybe its just me?) don't know what formulas to use.

 

[PAllen]

I agree if the infalling mass is still 1 kg, but after falling from a speed of 0 to 95%c, would the infalling mass still be 1 kg if you are measuring local to the initial stationary mass at the moment of collision?

The question I am really trying to get to: What does the distant observer see as the apparent speed and mass of the resulting object from the collision. The distant observer will see a certain speed and can infer a mass based on additional collisions. You see lots of correct animations for particle colliders, but you don't see many gravity animations of this since the computer programmers like me (maybe its just me?) don't know what formulas to use.

If it was 1kg at the point of dropping, then, for the observer I described, they would still use exactly 1 kg for its rest mass in applying normal collision formulas. It would be indistinguishable from a 1 kg object that was locally accelerated to 95% c. This is true as long as there is some operational definition of 1 Kg, and the same definition was used to prepare the object before the drop as was used by the lower observer to prepare local objects of 1 kg.

[edit: I didn't look much at the second part of your post. The thing is, that there is a as sense in which the distant observer would see the 1 kg falling body as having lower rest mass at the bottom; however, they would also see 1 Kg objects prepared at the bottom as being similarly 'lighter' than theirs (prepared using the same operational procedure). The result is that what I wrote above is also true. This is one thing JesseM has been trying to get across - you have to be really careful about who is measuring mass, and what their procedure is. There is no simple, universal definition of mass in GR.

 

[edguy99]

If it was 1kg at the point of dropping, then, for the observer I described, they would still use exactly 1 kg for its rest mass in applying normal collision formulas. It would be indistinguishable from a 1 kg object that was locally accelerated to 95% c. This is true as long as there is some operational definition of 1 Kg, and the same definition was used to prepare the object before the drop as was used by the lower observer to prepare local objects of 1 kg.

[edit: I didn't look much at the second part of your post. The thing is, that there is a as sense in which the distant observer would see the 1 kg falling body as having lower rest mass at the bottom; however, they would also see 1 Kg objects prepared at the bottom as being similarly 'lighter' than theirs (prepared using the same operational procedure). The result is that what I wrote above is also true. This is one thing JesseM has been trying to get across - you have to be really careful about who is measuring mass, and what their procedure is. There is no simple, universal definition of mass in GR.

How it was prepared is important, I had not thought of that. As the object gets lower I reduce the mass. Any idea how that is calculated?

 

[pervect]

The mass may or may not get lower depending on what sort of mass you mean. It's defintely premature to say that "as the object gets lower, I reduce the mass.", at least without some discussion of what you mean by mass.

In general, you'd be better of saying that the energy gets lower, rather than the mass gets lower - it's more in line with current usage, and less likely to be misunderstood.

 

[edguy99]

The mass may or may not get lower depending on what sort of mass you mean. It's defintely premature to say that "as the object gets lower, I reduce the mass.", at least without some discussion of what you mean by mass.

In general, you'd be better of saying that the energy gets lower, rather than the mass gets lower - it's more in line with current usage, and less likely to be misunderstood.

Videos of solar system gravity animations http://www.animatedphysics.com/videos/planets.htm" are based on each particle having:
1/ a position in space (px, py, pz)
2/ a movement in space (mpx, mpy, mpz) - or how much did it move in the last time step
3/ a mass and a radius

Between each frame, you:
1/ assume movement continues
2/ add in the force called gravity from Newtons inverse square law (1/r^2) given the mass and position of particles.

Some of these simulations are over 8 digits in accuracy and you start to wonder if you could model the precession of Mercury around the sun. Unfortunately, you quickly realize that anything modeled with the inverse square law will be an elipse with no precession. The easiest way around this and to be consistent with relativity is to add in a 1/r^3 force between bodies (ignoring 1/r^4+1/r^5... for the moment). This gets you the precession of whatever you want as shown in the first 3 videos http://www.animatedphysics.com/videos/precession.htm" .

These calculations are added:
1/ gravity includes a 1/r^3 component
2/ mass is added or reduced depending on if the particle is absorbing or losing energy.

The middle row http://www.animatedphysics.com/videos/precession.htm" shows attraction and repulsion modeled with frame dragging and the bottem row represents mass falling into a small spinning black hole under the kerr metric. It is interesting that all the mass is pulled out of the object as energy long before it hits the event horizion (which seems odd?).

I think I agree with you on the use of the word energy. In fact, in my mind I think of an electron as being .5MeV/c^2 and the proton as being 938MeV/c^2, but when talking about gravity, the term mass is more common I think. Also note that there is no setup here of a scale to "weigh" particles to get the mass. There is only one variable (or definition) of "weight" or "mass" or "energy", they are all the same. BUT, you can tell the "weight/mass/energy" of particles by crashing them into each other (using Newtons laws if things are slow, conservation of total energy and momentum if things are fast).